Year: 2019
Paper: 3
Question Number: 12
Course: LFM Stats And Pure
Section: Independent Events
There was a significant rise in the total entry this year with an increase of nearly 8.5% on 2018. One question was attempted by over 90%, two others were very popular, and three further questions were attempted by 60% or more. No question was generally avoided and even the least popular attracted more than 10% of the candidates. 88% restricted themselves to attempting no more than 7 questions, and only a handful, but not the very best, scored strongly attempting more than 7 questions.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1485.6
Banger Comparisons: 1
The set $S$ is the set of all integers from 1 to $n$. The set $T$ is the set of all distinct subsets of $S$, including the empty set $\emptyset$ and $S$ itself. Show that $T$ contains exactly $2^n$ sets.
The sets $A_1, A_2, \ldots, A_m$, which are not necessarily distinct, are chosen randomly and independently from $T$, and for each $k$ $(1 \leq k \leq m)$, the set $A_k$ is equally likely to be any of the sets in $T$.
\begin{questionparts}
\item Write down the value of $P(1 \in A_1)$.
\item By considering each integer separately, show that $P(A_1 \cap A_2 = \emptyset) = \left(\frac{3}{4}\right)^n$.
Find $P(A_1 \cap A_2 \cap A_3 = \emptyset)$ and $P(A_1 \cap A_2 \cap \cdots \cap A_m = \emptyset)$.
\item Find $P(A_1 \subseteq A_2)$, $P(A_1 \subseteq A_2 \subseteq A_3)$ and $P(A_1 \subseteq A_2 \subseteq \cdots \subseteq A_m)$.
\end{questionparts}For every element in $S$ we can choose whether or not it appears in a subset of $S$, therefore there are $2^n$ choices so $2^n$ distinct subsets.
\begin{questionparts}
\item $\mathbb{P}(1 \in A_1) = \frac12$ (since $1$ is in exactly half the subsets)
\item $\,$ \begin{align*}
&& \mathbb{P}(A_1 \cap A_2 = \emptyset) &= \mathbb{P}(i \not \in (A_1 \cap A_2) \forall i) \\
&&&= \prod_{i=1}^n \left ( 1-\mathbb{P}(i \in A_1 \cap A_2) \right) \\
&&&= \prod_{i=1}^n \left ( 1-\mathbb{P}(i \in A_1)\mathbb{P}(i \in \cap A_2) \right) \\
&&&= \prod_{i=1}^n \left ( 1-\frac12 \cdot \frac12\right) \\
&&&= \left (\frac34 \right)^n
\end{align*}
\item $\,$ \begin{align*}
&& \mathbb{P}(A_1 \cap A_2 \cap A_3 = \emptyset) &= \mathbb{P}(i \not \in (A_1 \cap A_2 \cap A_3) \forall i) \\
&&&= \prod_{i=1}^n \left ( 1-\mathbb{P}(i \in A_1 \cap A_2 \cap A_3) \right) \\
&&&= \prod_{i=1}^n \left ( 1-\mathbb{P}(i \in A_1)\mathbb{P}(i \in \cap A_2))\mathbb{P}(i \in \cap A_3) \right) \\
&&&= \prod_{i=1}^n \left ( 1-\frac12 \cdot \frac12 \cdot \frac12\right) \\
&&&= \left (\frac78 \right)^n
\end{align*}
Similarly, $\displaystyle \mathbb{P}(A_1 \cap A_2 \cap \cdots \cap A_m = \emptyset) = \left ( \frac{2^m-1}{2^m} \right)^n$
\item $\,$ \begin{align*}
&& \mathbb{P}(A_1 \subseteq A_2) &= \mathbb{P}(A_1 \cap A_2^c = \emptyset) \\
&&&= \left (\frac34 \right)^n \\
\\
&& \mathbb{P}(A_1 \subseteq A_2 \subseteq A_3) &= \prod_{i=1}^n \mathbb{P}(\text{once }i\text{ appears it keeps appearing}) \\
&&&= \prod_{i=1}^n \frac{\#\{(0,0,0), (0,0,1), (0,1,1), (1,1,1) \}}{2^3} \\
&&&= \prod_{i=1}^n \frac{4}{8} \\
&&&= \frac{1}{2^n} \\
\\
&& \mathbb{P}(A_1 \subseteq A_2 \subseteq \cdots \subseteq A_m) &= \prod_{i=1}^n \frac{m+1}{2^m} \\
&&&= \left ( \frac{m+1}{2^m} \right)^n
\end{align*}
\end{questionparts}
A quarter of the candidates attempted this scoring on average about half marks, making it the second most successfully attempted question. Candidates using the approaches suggested by the question tended to make good progress. Many produced correct solutions using various combinatorial arguments, some of which were easier to generalise than others and not all were well‐explained. In part (ii), a few candidates used arguments not permitted by the wording of the question. In part (iii), some obtained the first result by the nice method P(A₁⊆A₂) = P(A₁∩A₂'=∅) which is of course the first result of (ii) but this was not easy to generalise. Several incorrectly assumed P(A₁⊆A₂⊆A₃) = P(A₁⊆A₂)P(A₂⊆A₃) "by independence".